Tak and Hotels II

题目描述

N hotels are located on a straight line. The coordinate of the i-th hotel (1≤i≤N) is xi.
Tak the traveler has the following two personal principles:
He never travels a distance of more than L in a single day.
He never sleeps in the open. That is, he must stay at a hotel at the end of a day.
You are given Q queries. The j-th (1≤j≤Q) query is described by two distinct integers aj and bj. For each query, find the minimum number of days that Tak needs to travel from the aj-th hotel to the bj-th hotel following his principles. It is guaranteed that he can always travel from the aj-th hotel to the bj-th hotel, in any given input.

Constraints
2≤N≤10⁵
1≤L≤10⁹
1≤Q≤10⁵
1≤xi<x2<…<xN≤10⁹
x_i+1−x_i≤L
1≤aj,bj≤N
aj≠bj
N,L,Q,xi,aj,bj are integers.
Partial Score
200 points will be awarded for passing the test set satisfying N≤10³ and Q≤10³.

输入

The input is given from Standard Input in the following format:
N
x1 x2 … xN
L
Q
a1 b1
a2 b2
:
aQ bQ

输出

Print Q lines. The j-th line (1≤j≤Q) should contain the minimum number of days that Tak needs to travel from the aj-th hotel to the bj-th hotel.

样例输入

9
1 3 6 13 15 18 19 29 31
10
4
1 8
7 3
6 7
8 5

样例输出

4
2
1
2

提示

For the 1-st query, he can travel from the 1-st hotel to the 8-th hotel in 4 days, as follows:
Day 1: Travel from the 1-st hotel to the 2-nd hotel. The distance traveled is 2.
Day 2: Travel from the 2-nd hotel to the 4-th hotel. The distance traveled is 10.
Day 3: Travel from the 4-th hotel to the 7-th hotel. The distance traveled is 6.
Day 4: Travel from the 7-th hotel to the 8-th hotel. The distance traveled is 10.

学到了新东西......倍增 LCA

然后迷迷糊糊，有点一知半解

https://blog.csdn.net/lw277232240/article/details/72870644

这代码也是别人的

f[x][y]表示第x个点2^y次方的天数能到的最远点。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+10;
ll a[maxn];
ll f[maxn][35];
int main(){
    int n,step,q,i,j,p;
    scanf("%d",&n);
    for (i=0;i<n;i++)  scanf("%lld",&a[i]);
    scanf("%d%d",&step,&q);
    for (i=0;i<n;i++)  {
        p=upper_bound(a,a+n,a[i]+step)-a;//返回大于val的第一个位置
        f[i+1][0]=p;
    }
    for (int j=1;j<=30;j++)
        for (int i=1;i<=n;i++)
            f[i][j]=f[f[i][j-1]][j-1];
    int l,r;
    while (q--){
        scanf("%d%d",&l,&r);
        if (l>r) swap(l,r);
        ll ans=0;
        for (int i=30;i>=0;i--){
            if (f[l][i]<r){
                ans+=(ll)(1<<i);
                l=f[l][i];
            }
        }
        printf("%lld\n",ans+1);
    }
    return 0;
}